Prove that assuming $f:S\rightarrow T$, $f$ is a bijection iff there is $g:T\rightarrow S$ such that $f\circ g$ and $g\circ f$ are identity maps

Question

I'm trying to prove the following: Let $S$ and $T$ be sets and $f: S \rightarrow T$. Show that $f$ is a bijection iff there is a mapping $g: T \rightarrow S$ such that $f \circ g$ and $g \circ f$ are the identity mappings on $T$ and $S$, respectively.

Here's my proof:
Forward implication:Assume $f$ is a bijection. Then it's invertible and we have $f \circ f^{-1} = \text{id}_T$ and $f^{-1} \circ f = \text{id}_S$.

Reverse implication: Assume there is a mapping $g: T \rightarrow S$ such that $f \circ g = \text{id}_T$ and $g \circ f = \text{id}_S$. Assume $f$ is not injective. So $\exists x, y \in S \text{ with } f(x) = f(y) \text{ and } x \ne y$. This means $g(f(x)) = g(f(y))$, which is a contradiction since $x \ne y$ and $g \circ f = \text{id}_S$.

Now assume $f$ is not surjective, which means $\exists b \in T \ \forall a \in S \text{ such that } f(a) \ne b$. This means $f$ is not defined at $g(b)$, so $f \circ g \ne \text{id}_T$, which is a contradiction.

We've shown that $f$ is injective and surjective, so it is a bijection. $\blacksquare$

My questions are: Is this proof correct and satisfactory? Is there a better/simpler way to go about it?
Thanks in advance

Answer 1

Reverse implication seems right, but we have a problem with the forward implication, let's discuss it.

Since you have done the reverse implication right (or so we have assumed) let's just focus on the forward implication, in other words we are looking at the following statement

Let $S$ and $T$ be sets and $f:S\rightarrow T$ is a function. Show that if $f$ is a bijection then there exists a function $g:T\rightarrow S$ such that $g\circ f = \text{id}_S$ and $f\circ g = \text{id}_T$.

first ask yourself what is it that we are asked to prove, well in this case we are asked to prove that if $f$ is a bijection then $f$ is invertible, that is because the section "there exists a function $g:T\rightarrow S$ such that $g\circ f = \text{id}_S$ and $f\circ g = \text{id}_T$." means that $f$ is invertible and $g$ is the inverse.

your proof used the fact that if $f$ is bijective then it is invertible in the attempt of proving that if $f$ is bijective then it is invertible (circular reasoning) which makes it invalid.

to avoid making this common mistake (which is never completely avoidable) ask yourself what do I have and where can I go from there, in this case you have $f$ and you have the fact that it is bijective.

I can't pull $f^{-1}$ out of the hat as I don't have it, and I have nothing that leads to it.

Second ask yourself what solves the problem, in this case it is the finding of $g$, if I can find a function $g$ (explicitly) then done.

one way to do this (which happens to be the way that is going to work) is to construct the object that you wish to find using what you have (in this case we have a bijection $f$, and we are trying to construct the function $g$ such that it satisfies the conditions we want it to satisfy)

you can easily prove that the relation given by (if $f(x)=y$ then $g(y)=x$) is a function, furthermore you can easily prove that it satisfies the conditions.

you might say but in this case $g$ is just the inverse function of $f$, that is correct but in your proof it looked like you pulled $f^{-1}$ out of the hat, but in my proof we carefully constructed $g$ using what we have.

I follow this way of reasoning to this day and it made me better that is why I wanted to showcase the entire thought process. thank you for reading